/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Positive charge \(Q\) is distrib... [FREE SOLUTION] | 91Ó°ÊÓ

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Positive charge \(Q\) is distributed uniformly over each of two spherical volumes with radius \(R .\) One sphere of charge is centered at the origin and the other at \(x=2 R\) (Fig. P22.63). Find the magnitude and direction of the net electric field due to these two distributions of charge at the following points on the \(x\) -axis: (a) \(x=0 ;\) (b) \(x=R / 2 ;(\mathrm{c}) x=R ;(\mathrm{d}) x=3 R\)

Short Answer

Expert verified
At (a)x=0: \( \frac{kQ}{4R^2} \) (positive x); (b)x=R/2: \( \frac{4kQ}{9R^2} \) (positive x); (c)x=R: 0; (d)x=3R: \( \frac{8kQ}{9R^2} \) (negative x).

Step by step solution

01

Understanding the Setup

Two spheres are centered on the x-axis, the first sphere is centered at the origin (0,0) and the second sphere is at point (2R,0). Each sphere has a uniformly distributed charge, Q. Calculate the electric field at different points along the x-axis due to these spheres.
02

Determine the Contribution of Each Sphere

For a point outside a uniformly charged sphere, the sphere behaves like a point charge located at its center. Hence, the electric field caused by a sphere at a distance, r, from its center is given by \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant.
03

Calculate the Electric Field at x=0

At the origin (x=0), the field by sphere 1 is zero as the point is inside it. The electric field by sphere 2 at x=0 is directed towards the positive x-direction and is given by \( E_2 = \frac{kQ}{(2R)^2} = \frac{kQ}{4R^2} \).
04

Calculate the Electric Field at x=R/2

At \( x=\frac{R}{2} \), the point is inside sphere 1 so the electric field due to sphere 1 is zero. For sphere 2, the distance is \( 2R - \frac{R}{2} = \frac{3R}{2} \). The field due to sphere 2 is \( E_2 = \frac{kQ}{(\frac{3R}{2})^2} = \frac{4kQ}{9R^2} \). Direction: positive x-axis.
05

Calculate the Electric Field at x=R

At \( x=R \), for sphere 1, \( r = R \), so \( E_1 = \frac{kQ}{R^2} \); direction is negative x-axis. For sphere 2, \( r = R \), \( E_2 = \frac{kQ}{R^2} \); direction is positive x-axis. Net field is zero as they are equal and opposite.
06

Calculate the Electric Field at x=3R

At \( x=3R \), for sphere 1, \( r = 3R \), so \( E_1 = \frac{kQ}{(3R)^2} \) in positive x-direction. For sphere 2, \( r = R \), \( E_2 = \frac{kQ}{R^2} \) in negative x-direction. Net \( E = E_2 - E_1 = \frac{kQ}{R^2} - \frac{kQ}{9R^2} = \frac{8kQ}{9R^2} \) in negative x-direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is the essential formula used to calculate the electric force and electric field due to point charges. It states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The mathematical expression is given by:
\[ F = \frac{k |q_1 q_2|}{r^2} \]where:
  • \( F \) is the magnitude of the force between the charges,
  • \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
  • \( r \) is the distance between the centers of the two charges.
When it comes to the electric field \( E \), produced by a charge \( Q \), Coulomb's Law adapts to:
\[ E = \frac{kQ}{r^2} \]Here, \( E \) represents the electric field at a point at a distance \( r \) from the charge. This field is the force per unit charge and acts along the line connecting the charge and the point of interest.
Uniform Charge Distribution
When a charge is uniformly distributed, it means that the charge is spread evenly over a surface or throughout a volume. In the context of a spherical volume, it suggests that the charge is distributed so evenly that any chosen section of the sphere holds the same charge density. The charge density \( \rho \) (charge per unit volume) for a volume charge is calculated by:
\[ \rho = \frac{Q}{V} \]where \( V \) is the volume of the sphere. For a sphere, this volume is given by \( V = \frac{4}{3} \pi R^3 \). Therefore, if a charge \( Q \) is uniformly distributed across a sphere of radius \( R \), the charge density becomes:
\[ \rho = \frac{Q}{\frac{4}{3} \pi R^3} \]Understanding this helps in deducing the electric field both inside and outside the sphere. Internally, within a charged sphere, the symmetry results in a net field of zero at any point inside, since contributions from all parts of the charge cancel out. Externally, the sphere behaves as if all its charge were concentrated at its center.
Spherical Charge Distribution
A spherical charge distribution refers to a setup where a charge is spread out evenly across the volume of a sphere. This creates certain symmetrical properties in how the electric field behaves both inside and outside the sphere.
  • **Inside the Sphere:** The electric field inside a uniformly charged sphere is zero. This is because the charges are symmetrically placed around any given interior point, leading to uniform cancellation of forces from all directions.
  • **Outside the Sphere:** For points beyond the surface of a charged sphere, the sphere behaves like a single point charge located at its center. This means the electric field at any external point at a distance \( r \) from the center is calculated using Coulomb's Law as if the sphere were merely a point charge:
\[ E = \frac{kQ}{r^2} \]This simplification is extremely beneficial when calculating fields along axes or in certain symmetric arrangements like in the given exercise.
Electric Field Direction
The direction of an electric field is crucial in understanding how charges influence each other. By convention, the electric field direction is the direction that a positive test charge would move if placed in the field. For a positive charge, the electric field points radially outward, while for a negative charge, it points inward.
A simple understanding involves:
  • **Positive Charges:** Generate an outward field away from the charge.
  • **Negative Charges:** Produce an inward field towards the charge.
  • **Influence of Multiple Charges:** When dealing with multiple charges, as in the exercise with two spheres, the total electric field at a point is the vector sum of the fields due to each charge distribution.
In the specific example of the spheres with charge \( Q \), each sphere contributes an electric field in a certain direction:
  • At points on the x-axis, the field's direction is determined by comparing which sphere is closer or further, considering the signs and magnitudes of each contributing field.
  • The net direction is the result of additive or subtractive interactions based on the field lines from each charge distribution.

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Most popular questions from this chapter

How many excess electrons must be added to an isolated spherical conductor \(32.0 \mathrm{~cm}\) in diameter to produce an electric field of \(1150 \mathrm{~N} / \mathrm{C}\) just outside the surface?

A square insulating sheet 80.0 cm on a side is held horizontally. The sheet has 7.50 nC of charge spread uniformly over its area. (a) Calculate the electric field at a point 0.100 mm above the center of the sheet. (b) Estimate the electric field at a point 100 m above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?

A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho .\) (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\) . What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r=R .\) (d) Graph the electric-field magnitude as a function of \(r\) from \(r=0\) to \(r=3 R .\)

You measure an electric field of \(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\) at a distance of 0.150 \(\mathrm{m}\) from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?

Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius \(R .\) Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located (a) a distance \(r > R\) from the center of the shell (outside the shell) and (b) a distance \(r < R\) from the center of the shell (inside the shell).

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